Question

The sum of the real roots of the equation $(7+4\sqrt{3}{)}^{x^2-8}+(7-4\sqrt{3}{)}^{x^2-8}=14$ is

Answer 1

$(7+4\sqrt{3}{)}^{x^2-8}+(7-4\sqrt{3}{)}^{x^2-8}=14$

$(7+4\sqrt{3})\times (7-4\sqrt{3})=49-16\times 3=1$
$\Rightarrow 7-4\sqrt{3}=\frac{1}{7+4\sqrt{3}}$
Assuming $(7+4\sqrt{3}{)}^{x^2-8}=t$, we get
$t+\frac{1}{t}=14\Rightarrow t^2-14t+1=0\Rightarrow t=\frac{14\pm \sqrt{196-4}}{2}\Rightarrow t=7\pm 4\sqrt{3}$

Now, $(7+4\sqrt{3}{)}^{x^2-8}=7\pm 4\sqrt{3}$
$\Rightarrow x^2-8=\pm 1\Rightarrow x^2=7,9\Rightarrow x=\pm \sqrt{7},\pm 3$

Hence, the sum of real roots $=0$